CAB 3023: PROCESS PLANT DESIGNMaterial and Energy Balances

3.2.4 Distillation ModelsCalculating Linear Split Fractions

The previous chapters introduced a systematic strategy for generating candidate flowsheets. This chapter deals with the development of simple, fast, and useful methods for evaluating the

behavior of a candidate flowsheet. Often the rules involved in this process lead to the elimination of several undesirable alternatives. The remaining alternatives, however, require a more detailed evaluation and this task forms the

basis of the next three chapters. In particular, this chapter develops simple strategies for obtaining mass and energy balances for a

candidate flowsheet. This task is one of the most necessary and the most time-consuming for flowsheet evaluation. Still, with the simplifications introduced in this chapter, the mass and energy balances can be

calculated quickly and a great deal of insight is gained in the process. Nevertheless, the simplifications in this chapter lead to inaccuracies in the final flowsheet that need

to be corrected with more detailed models (for example, by using a process simulator). These will be discussed in later chapters.

3.1. INTRODUCTION

To evaluate the conceptual flowsheet presented in the previous chapters, we need to consider the detailed and time-consuming task of heat and mass balances.

This precedes the later tasks of plant equipment sizing and economic evaluation.

For more detailed mass and energy balances, on the other hand, there are many computer programs or process simulators that perform these tasks in a more rigorous way.

Typically, a candidate flowsheet model can be defined as a large set of nonlinear equations describing: the connectivity of the units of the flowsheet through process streams; the specific equations for each unit; these usually deal with internal mass and energy balances

as well as equilibrium relationships; underlying physical property relationships that define enthalpies, equilibrium constants, and

other transport and thermodynamic properties.

Taken together, these equations can number in the many thousands. To deal with them directly, two methods for flowsheet simulation, the modular and equation-

oriented modes, have been developed and incorporated into engineering practice. While a complete description of these modes is deferred to a later chapter, a little background

is also useful here.

3.2. DEVELOPING UNIT MODELS FOR LINEAR MASS BALANCES

Once temperature and pressure are fixed in the input and output streams, we can develop a set of equations for each process unit and thereby solve the entire flowsheet with these equations.

Thus, our overall strategy will be:1. fix temperature and pressure for all process streams;

2. approximate each unit with split fractions representing outlet molar flows linearly related to the inlet molar flows;

3. combine the linear equations and solve the overall mass balance;4. recalculate stream temperatures and pressures from equilibrium relationships;5. if there are no large changes in temperature and pressure, go to Step 6, else, go to Step 1;6. given all temperatures and pressures, perform the energy balance and evaluate heat duties.

To follow this decomposition, we assume that all vapor and liquid streams have ideal equilibrium relationship (particularly in step 2) and that, unless otherwise stated, all streams are at saturated conditions.

With these assumptions, physical properties can be calculated from standard handbook data. In this text, we rely on Reid, Prausnitz, and Poling (1987) as our data source. The advantages of this approach are that calculations are very easy to set up and solve with few

iterations (usually no more than two (required for convergence of a preliminary design.

Consider the flowsheet shown in Figure 3.1, with the units shown as rectangles connected by inlet (input) and outlet (output) streams.

In this section, we construct linear model approximations for the following units: mixer, splitter, reactor, flash, distillation column, absorber, stripper.

The above list contains a comprehensive set of mass balance units and in the next section, we will show how to put the flowsheet in Figure 3.1 together with them.

Additional information on the shortcut separation units can also be found in Douglas (1988) and Perry et al. (1984).

To construct the linear unit models, we label the stream vector of molar flows I,j as the jth output stream of unit I.

is the flowrate of component k in this stream.

Also, if there is only one outlet stream in unit I, the j subscript is suppressed. Note that with this notation, we express stream composition in terms of molar flows instead of

mole fractions, as this preserves the linearity of the equations, For example, for unit 2 in Figure 3.2, 2,2,CH4 refers to the molar flowrate of CH4 in the second

effluent stream.

3.2.1. Linear Mass Balances for Simple Units

Mixer Unit

Figure 3.3. Mixer unit

The mixer unit simply sums all of the inlet streams as a single output stream with the following mass balance equations.

Given upstream units i1, i2, …, iN that feed into the mixer with the j1th outlet from unit i1, the j2th outlet from unit i1, etc., for component k, M is written as:

(13)

Splitter Unit

Figure 3.4. Splitter unit

The splitter unit divides a given feed stream into specified fractions j for each output stream j. Note that all output streams have the same composition as the feed stream. Thus, for NS output streams, we have (NS – 1) degrees of freedom in choosing j and write the

equations:

Reactor (Fixed Conversion Model)

For linear mass balances, we assume that the reactor model can be simplified by specifying the molar conversion of the NR parallel reactions in advance (Figure 3.5).

As a result, the mass balance equations remain linear and relatively easy to solve. For each reaction r, we define a limiting component l(r), and normalized stoichiometric

coefficients , r = 1, …, NR for each component k, in which the coefficients Ck,r

appear in the specified reactions. We also adopt the convention:

Defining the fraction converted per pass based on limiting reactant as r, r = 1, …, NR gives us:

The equations for the fixed conversion reactor model are best illustrated by example.

3.2.2. Calculation of Flash Units—the “Building Block” Unit in Process Flowsheets

This calculation is the most fundamental and important one in a flowsheet.

Aside from the physical separation unit itself, it is the building block for deriving linear models for equilibrium-staged separations such as distillation and absorption.

These calculation procedures will also be used later for setting pressures and temperatures around the flowsheet.

We first consider the simple phase separation unit described in Figure 3.6, as well as a number of calculation procedures for this unit.

Figure 3.6. Liquid–vapor flash unit

To develop the flash model, we first define an overhead split fraction k = vk/fk for each of the ncomp components k.

We further identify component n as a key component (for which a given recovery can be obtained) and also define vapor fraction = V/F for specified vaporization of the feed.

As specifications, the variables n,m, , P, T, and Q (heat supplied to flash unit), can be specified. If we now write the equations for the flash unit:

We find that for a specified feed, the (number of variables) – (number of equations) = 2 degrees of freedom.

This means that we can completely specify the condition of the flash unit if we select two of the variables.

Since we have not yet considered energy balances, we defer specifications on Q and now consider the following cases: Case 1: n specified (key component overhead recovery) and T or P specified; Case 2: T and P specified (isothermal flash); Case 3: specified and T or P specified.

The first case is very useful for the shortcut methods in this chapter, but is not used for more detailed models.

Cases 2 and 3 are needed for analyzing design and operating conditions.

We now consider some approximations for vapor–liquid phase equilibrium (VLE). Equating the mixture fugacities in each phase leads to a reasonably general expression at low to

moderate pressures:

where k is a vapor fugacity coefficient, k is the liquid activity coefficient, and is the pure

component fugacity. For process calculations, it is often convenient to represent the equilibrium relation as

with the K value given by .

For our shortcut calculations, we assume ideal behavior which leads to the following assumptions: k = 1, k = 1,

(vapor pressure)

Antoine equation for vapor pressure:

Where the Antoine equation is a representative correlation with coefficients that can be found, for example, in Reid et al. (1987).

These assumptions lead to Raoult’s Law:

or more simply:

With respect to key components, we can now define a relative volatility:

which, for ideal systems, is independent of P and is much less sensitive to T than Kk is. Note that component k can be nonvolatile, in which case k/n 0. On the other hand, if component k is noncondensible, k/n . We can now rederive and simplify the flash equations. Let:

We now reintroduce the split fractions and define:

and

Substituting these definitions into the above equations gives us:

as well as: at equilibrium.

Rearranging this expression gives:

and we have now defined the recovery of each component in terms of the key component recovery. Note also that the limiting cases of nonvolatile (k/n 0, k 0) and noncondensible (k/n , k

1) components are also observed.

With specification of the key component recovery, an additional specification is still required (because recall that we require two degrees of freedom).

Implicit in the above expression is that a correct value of temperature (T) was known in advance in order to calculate the relative volatilities,

Given that we have specified T or P, how do we calculate the corresponding value of P or T? Moreover, if we have specified T or P directly, how do we use the above equation to determine the

corresponding key component recovery? Here we need to consider a bubble (or dew point) equation that also needs to be satisfied at

equilibrium. At the bubble point1 (for the saturated liquid effluent stream), we have:

1 bubble point (of a mixture of liquids at a given pressure): the temperature at which the first vapor bubble appears when the mixture is heated

or in terms of relative volatilities:

where is defined as an average relative volatility. Using this definition allows us to redefine the K-value as:

which forms a simplified bubble point equation. For T fixed and P unknown, we can calculate a value of P directly from:

On the other hand, for P fixed and T unknown, the value of T can be calculated approximately from:

To reduce approximation errors, we choose the index k to be the most abundant component in the liquid phase.With the above equations, we can now develop the following algorithms for the three most commonly specified flash problems.

Case 1: n and P (or T) fixed1. For a specified n and P (or T), guess T (or P).2. Calculate Kk, k/n at specified T.

3. Evaluate for each component k.

4. Reconstruct a mass balance and calculate mole fractions:

5. For T fixed, .

For P fixed, solve T from .

Case 2: T and P fixed1. For a specified T and P, pick a component n and guess n.2. Follow steps 2, 3, and 4 for the algorithm of Case 1.

3. If the bubble point equation is satisfied: , stop. Otherwise, reguess n, and go to step

3. (Simple iterative methods such as the secant algorithm learn in the course on numerical methods, can be used to obtain convergence for n.)

Case 3: and P (or T) fixed

1. For a specified = V/F and P (or T), guess T (or P), calculate k/n, Kk, and define .

2. Define .

3. Then follow steps 3 and 4 of the previous algorithm.

4. If the bubble point equation is satisfied: , stop. Otherwise, reguess T (or P), and go

back to step 1. (Simple iterative methods such as the secant algorithm learn in the course on numerical methods, can be used to obtain convergence for n.)

These algorithms have been stated very concisely.Each of these algorithms will be illustrated by the following long example.

EXAMPLE 3.2: Flash CalculationConsider the mixture with the components, flowrates, boiling points, and Antoine coefficients given in the following table:

Antoine coefficient

Component k Flowrate fk

(kmol/h)Boiling point (K)

Ak Bk Ck

Benzene 30 353 15.9008 2788.51 52.34Toluene 50 383 16.0137 3096.52 53.67o-xylene 40 418 16.1156 3395.57 59.44

Here we choose toluene as the key component (n = 2) because of its intermediate volatility.

Case 1: Fixed 2 = 0.9 and P = 1 bar

If we assume that k/n remains constant over the temperature range, we can do a direct calculation without iteration.

1. Specify 2 = 0.9,P = 1 bar, and guess T = 390 K.

2. Calculate relative volatilies (same as above problem).

1/2 = 2.3053/2 = 0.381

3. Calculate recoveries of nonkey components:1 = 0.9542 = 0.774

4. Solve for material balance and evaluate mole fractionsv1 = 28.62 l1 = 1.38 x1 = 0.089v1 = 28.62 l1 = 1.38 x1 = 0.089

5. Evaluate bubble point equation:

but T(P0 = 380 mmHg) = 393 K (estimate of T is close enough).

Case 2: Flash calculation at 1 bar and 390 K

Case 3: Vapor fraction = 0.8 and P = 1 bar

1. = 0.8, P = 1 bar, guess T = 390 K2. Evaluate K values, relative volatilities, and key component recovery3. Evaluate nonkey component recoveries4. Solve material balances and evaluate mole fractions

BUBBLE AND DEW POINT CALCULATIONS

The algorithm presented above allow rapid calculation of flash separators. However, in the limiting cases of the bubble point ( = 0) or the dewpoint ( = 1), these algorithms

can be further simplified and are given in Figures 3.7 and 3.8.

Bubble point algorithm: = 0 (saturated liquid)

Figure 3.7.

Here, k = 0, lk = fk, and xk = zk.

Case 1: For P fixed, calculate T directly from .

Case 2: For T fixed, calculate P from .

In both Cases 1 and 2, n is chosen as the most abundant component.

Dew point algorithm: = 1 (saturated vapor)

Here, k = 1, vk = fk, and yk = zk. For this case, we derive a dew point equation based on: yk = zk.

Here:

Select as k = n as the most abundant vapor component.

Then and

for T fixed:

for P fixed:

Again, a key assumption for this last equation is that k/n remains fairly constant with temperature.

Upper Limits of Pressure and Temperature in Vapor–Liquid Equilibrium

Of course, the above simplified flash calculations (as with more detailed calculations) cannot be applied at or above the critical region.

At the critical point, we have equal densities for the vapor and liquid phases. If we examine the phase diagram for mixtures, illustrated in Figure 3.9, we note some unusual

behavior not described by the flash algorithms. For example, in the region of isobaric retrograde condensation, above the critical pressure,

increasing temperature will lead to liquefaction. Similarly, for the region of isothermal retrograde condensation, above the critical temperature, and

increase in pressure will lead to increased vaporization.

Calculations in the neighborhood of the critical point still remain important challenges for detailed phase equilibrium algorithms.

For the purpose of our simplified design calculations, we will simply avoid critical regions by using the following guideline to test the existence of a liquid phase.

Here, we define a pseudocritical temperature for a mixture as: is the critical

temperature of component k. Here, we use liquid mol fractions because these give more realistic estimate of critical temperature

for mixtures.

Note: An immediate application of these upper limits is in specifying a pseudocomponent or a hypothetical component in a process simulator.

For an unknown component, by specifying (1) boiling point, and (2) T and P critical point, (3) density, we would be able to “characterize” that component.

Example 3.3Consider the mixture of the previous flash calculations. We would like to determine if the critical point of this mixture is above 392 K and 1 bar, the point at which we would like to flash this mixture. From handbook data, we have:

Tc (K) xk

Benzene 56.2 0.099Toluene 592O-xylene 630 0.560

and the mixture critical point is Tm,k = 610 K, well above the flash specification (392 K).

Example 3.4Consider the following H2/H2O system with mole fractions zk where we know a liquid phase exists at room temperature (300 K) and pressure (1 bar). From handbook values, we have the following properties:

Here, if we set water as the key component, we have 1/2 = 18 400. Assume that 2 = 0.01, we calculate 1 =0.994 and the following mass balance can be obtained as a rough guess:

3.2.4. Distillation Models

In this subsection, we establish split fractions based on simple shortcut methods for distillation. Distillation operations can be described as a cascade of equilibrium trays with each one solved as a

flash unit (Figure 3.10). The feed stream enters at an intermediate tray: at the bottom, liquid product is removed, a reboiler

vaporizes the liquid stream on the lowest stage, and counter-current liquid and vapor streams are set up in the distillation column.

Similarly, vapor leaving the top tray is condensed and overhead product is removed, with the remaining liquid returned or refluxed back to the top tray.

Detailed calculation of the tray-by-tray behavior of a distillation column will not be considered at this stage in the design, but will be deferred to a later chapter.

Instead, we will make a number of approximations using limiting column behavior (total reflux) in order to obtain linear mass balance models and relevant equipment parameters (i.e., temperatures and pressures of the column).

First, let us identify the degrees of freedom available for mass balance in a distillation column. For determining the mass balance, it turns out that if we know the overhead split fractions LK, HK

(where lk and hk refer to light and heavy key components, respectively) and the overhead column pressure, we have already fully specified the column equations.

So why are there only three degrees of freedom in a column mass balance, regardless of the number of distillation trays?

Intuitively, we can think of the top of the column, which further refines the light key with (1) lk, and (2) pressure at the top of column PT specified (as in a flash unit), and the bottom of the column, which further refines the heavy key with (3) hk, and (4) PB specified (as in a flash unit).

Thus, we have four specifications. But since PT + P = PB in which P is the column pressure drop, there are only three independent

degrees of freedom.

Calculating linear split fractions

To further specify a distillation column and derive the component recoveries in the linear mass balance equations, we classify five types of components:1. components lighter than the light key,2. light key component,3. components between keys (distributed components),4. heavy key component,5. components heavier than the heavy key.

As with the flash unit, we will assume that ideal behavior exists in our simplified column. From the flash unit, we will assume that ideal behavior exists in our simplified column. From the flash unit, we know k/n is independent of pressure, and we assume it is independent of

temperature in ideal situations. Moreover, in order to do the mass balance, we must know the split fractions of the distributed

components. After the mass balance, we also need to consider the number of trays and the temperatures in the

column. To find this, we use the Fenske (1932) equation for total reflux. This equation is easily derived and gives an approximate product distribution as well as an

estimate of the minimum number of trays. Consider the total reflux case shown in Figure 3.11 Here, we note that the feed and bottom streams are negligible compared to the total reflux flow and

can be ignored. Starting at the reboiler, we note from the mass balance of vapor and liquid streams above the

reboiler that . Also, from the equilibrium relation, at reboiler stage:

from:

At stage (N – 1), we can again write the equilibrium relation:

Similarly, at stage (N – 2), we have the relation:

Finally, since for every stage j, we can write:

Writing in terms of distillate and bottoms flowrates and defining split fractions for these, yields:

and with split fraction given by , we rearrange the above expression to yield:

If we have specified the light and heavy key recoveries, then the minimum number of stages is given directly by the Fenske equation:

Once we know Nm, all of the other component split fractions can be obtained simply by substituting k for LK in the above expressions.

With minor rearrangement, we have:

Note that this equation reduces to the split fraction for the flash unit when Nm = 1. Moreover, while the above equation applies to all components, we will simplify our analysis and

apply this equation to distributed components only. This follows because key component split fractions, LK and HK, will be specified close to one

and zero, respectively. Hence, for all but the distributed components, we can assume:

Component type Component recovery k

Lighter than light key 1 (k/HK > 1, as Nm , k = 1)Light key LK fixed (e.g., 0.99)Distributed component from equation for k

Heavy key HK fixed (e.g., 0.01)Heavier than heavy key 0 (k/HK < 1, as Nm , k = 0)

Once these split fractions are calculated, the linear mass balance for the distillation column is straightforward as depicted in Figure 3.12.

Setting Column Pressures and Temperatures

In addition to specifying recoveries of key components, we also need to set an appropriate pressure (or temperature) for the top of the column.

To do this, we first need to explore the constraints on these specifications. These are primarily dictated by the cooling water temperature (Tcw) in the condenser and the steam

supply temperature (Tst) in the reboiler. Consider Figure 3.13 with a total condenser and reboiler and with temperatures marked in different

column locations.

Since we know that the column pressure is lower at the top than the bottom, and that the more volatile (low boiling components) are also higher in concentration at the top, we note the following temperature relationships:

Tcw Tbub,C Tdew,C Tbub,R Tdew,R Tst

Column pressure can be selected so that the following constraints hold:1. select condenser pressure so that Tbub,C (Tcw (about 30C) + T (about 5 K)) Tbub,C ~ 310 K;2. select condenser pressure so that all bubble point temperatures are below the critical

temperature of a mixture, i.e.: ;

3. from the bubble point equation, we note Tbub increases with P and we prefer to choose P to be

above 1 atm. Thus, (below 1 atm, thicker vessel walls and additional

safety precautions are required to avoid air leaks and explosion hazards).

Special case of noncondensible and nonvolatile components

These constraints can be difficult to meet when we have both noncondensible (very low boiling) components or nonvolatile (very high boiling) components in the system.

One common way to still satisfy the above pressure restrictions is to consider partial condensers and reboilers for noncondensible and nonvolatile components, respectively.

Mass balances with these additional devices can be determined through an additional flash calculation.

Consider first the partial condenser shown in Figure 3.14.

Calculating the mass balance and temperatures around the partial condenser can be greatly simplified by noting that the product streams are at saturated liquid and vapor and can be obtained through a simple flash calculation, once the product flows and composition dk are specified.

From this, we note that the partial condenser can be represented schematically in Figure 3.15.

From this, a direct way to calculate the mass balance involves the following scheme:1. relate D to L through a predetermined reflux ratio (RR = L/D). RR can be determined from

using shortcut methods (Fenske – Underwood – Gilliland equations);2. to obtain Tcond, do a Case 3 flash calculation on the flash tank with P and

specified to get Tcond, yD, and xD. Note that the feed to this tank is given by dk. (The vapor fraction of the product can be specified, for example, by the fraction of noncondensible components in the product.

3. calculate L, V, and the dewpoint composition y1 in V from the mass balance equations:

4. to determine Tdew, perform a dew point calculation for V and P and y1 specified. These temperatures will be useful for sizing the condenser as well as for the energy balance.

Partial reboilers can also be analyzed in a simpler manner as shown in Figure 3.16.

Note that the dew point exiting the reboiler is the highest temperature in the column. To avoid excessively high temperatures, a partial reboiler effectively adds an extra equilibrium

stage. To calculate the difference in temperatures, the dewpoint temperature in a total reboiler is given

by:

where n is the most plentiful component and P = P + P. Here the composition yk is the same as the bottoms product and there is a large contribution in the

summation from high-boiling components. With a partial condenser, on the other hand, the composition yk is not as rich in these

components---both and Tdew are lower.

Similarly, the bubble point temperature for the reboiler product can be calculated from the bubble point equation.

Effect of Pressure on Separations

Before concluding this subsection, we note the effect of increasing pressure on the difficulty of the separation involved.

Under an ideal assumption, we see that is not directly affected by pressure.

However, it is indirectly related because bubble point temperature changes significantly with pressure and thus lead to significant differences in relative volatilities.

Therefore, as P becomes large, so do the partial pressures of the overhead product as well as the overhead temperature.

Moreover, for ideal systems,

EXAMPLE 3.5: DISTILLATION

Consider the separation of a benzene, toluene, and ortho-xylene mixture in which we would like to recovert 99% of the benzene in overhead and 99.5% of the o-xylene in the bottoms stream.We therefore choose benzene and o-xylene as light and heavy key components, respectively.Note the following data for the feed.

Component Flow (kgmol/h) K (386 K, 1 bar) LK/HK

Benzene (1) 20 2.52 6.209Toluene (2) 30 1.079 2.662o-xylene (3) 50 0.405 1.0

1 = 0.993 = 1 – 0.995 = 0.005

Step 1. Calculate minimum number of trays (at total reflux) using Fenske equation:

Step 2. Split fraction for the distributed component of toluene:

Step 3. Calculate mass balances from the split fractions

Step 4. Determine the pressure and temperature at the top of the column with a total condenser. Choose benzene as the most plentiful component and perform a bubble point equation.

Step 5. Calculate temperature using Antoine equation

Step 6. Check that the distillate temperature is higher than cooling water temperature (30C). this completes the calculation for Tbub,C.

Step 7. Calculate the overhead vapor temperature Tdew,C using dew point calculation:

Step 8. Repeat the same procedure for determining the mass balances, temperatures, and pressures for the bottom part of the column by choosing o-xylene as the most plentiful component.

3.2.4 GAS ABSORPTION WITH PLATE ABSORBERS

As with distillation, gas absoption can be modeled approximately as a cascade of equilibrium trays. The assumptions of equilibrium stages is weaker here, and as with distillation, we will seek to

correct this in the next chapter through the use of tray efficiencies. In this subsection, we will model two similar gas – liquid operations: aobsption and stripping. Absoption represents a vapor recovery operation in which a desired component is transferred from

a gas to the liquid phase