- 7.0 Summary—Objectives
- 7.1 Total Reflux: Fenske Equation
- 7.2 Minimum Reflux: Underwood Equations
- 7.3 Gilliland Correlation for Number of Stages at Finite Reflux Ratios
- References
- Problems
This chapter is from the book
This chapter is from the book
This chapter is from the book
7.1 Total Reflux: Fenske Equation
Fenske (1932) derived a rigorous solution for binary and multicomponent distillation at total reflux. The derivation assumes that the stages are equilibrium stages. A multicomponent distillation column with a total condenser and a partial reboiler operating at total reflux is shown in Figure 7-1. For any two components A and B, equilibrium in the partial reboiler R requires
Equation (7-1) is the definition of the relative volatility, Eq. (2-6a), applied to the reboiler. Material balances for components A and B around the reboiler are
and
However, at total reflux, B = 0, and LN = VR. Thus, the mass balances become
For a binary system, this naturally means that the operating line is the y = x line. Combining Eqs. (7-1) and (7-4),
If we now move up the column to stage N, the equilibrium equation is identical to Eq. (7-1), except it is written for stage N. The mass balances around stage N simplify to yA,N = xA,N−1 and yB,N = xB,N−1. Combining these equations, we have
Equations (7-5) and (7-6) can be combined to give
which relates the ratio of liquid mole fractions leaving stage N–1 to the ratio in the reboiler.
)
FIGURE 7-1. Total reflux column
Repeating this procedure of alternating between the operating and equilibrium equations, the result at the top stage is
If we define αAB as the geometric average relative volatility,
and note that reboiler and bottoms compositions are identical, Eq. (7-8a) becomes
Solving Eq. (7-9) for Nmin, we obtain
which is one form of the Fenske equation. Nmin is the number of equilibrium contacts including the partial reboiler required at total reflux. Although the derivation is for any two components, Eq. (7-10) is most accurate if it is written for the light key and the heavy key. Then we have
If the relative volatility is constant, Eq. (7-10) is exact.
An alternative form of the Fenske equation that is very convenient for multicomponent calculations is easily derived. First, rewrite Eq. (7-11a) as
DxA,dist is equal to the fractional recovery of A in the distillate multiplied by the amount of A in the feed, and BxB,bot is the fractional recovery of B in the bottoms multiplied by zB,
These equations are the multicomponent equivalent of Eqs. (3-5a) and (3-5c). Substituting these equations and the equations for DxB,dist and BxA,bot into Eq. (7-11b), and identifying A = LK and B = HK,
Equation (7-12) is in a convenient form for determining the number of stages for multicomponent systems.
For multicomponent systems, calculation with the Fenske equation is straightforward if fractional recoveries of the two keys, LK and HK, are specified. Equation (7-12) can be used directly to find Nmin. The relative volatility can be approximated by a geometric average. Once Nmin is known, the fractional recoveries of the NKs can be found by first writing Eq. (7-12) for an NK component and one of the key components. For example, if we replace LK with an NK component,
Solving for (FRNK,dist) the result is,
Remember that the order of subscripts on αAB is important.
If two mole fractions are specified, say xLK,bot and xHK,dist, the multicomponent calculation is more difficult. We cannot use the Fenske equation directly, but several alternatives are possible. If we can assume that all NKs are nondistributing, we can use the strategy used in Chapter 5 to do mass balances. Assume the NKs follow Eqs. (5-6a) to (5-8) and then calculate D and B from the summation equations, Eqs. (3-6a) and (3-6b). Once all distillate and bottoms compositions or values for DxD,i and BxB,i have been found, Eq. (7-11a) or (7-11b) can be used to find Nmin. Use the key components for this calculation. The assumption of nondistribution of the NKs can be checked with Eq. (7-13). If the original assumption is invalid, the calculated value of Nmin obtained for key components can be used to calculate the light non-key (LNK) and heavy non-key (HNK) compositions in distillate and bottoms. Then Eq. (7-11a) or (7-11b) is used again to obtain a more accurate estimate of Nmin.
If NKs distribute, a reasonable first guess for the distribution is required. This guess can be obtained by assuming that the distribution of NKs is the same at total reflux as it is at minimum reflux. The distribution at minimum reflux can be obtained from the Underwood equation (Case C) and is covered later.
The derivation up to this point has been for any number of components. If we now restrict ourselves to a binary system where xB = 1 − xA =1 − x, and Eq. (7-11a) becomes
where x = xA is the mole fraction of the more volatile component (MVC). The use of the Fenske equation for binary systems is quite straightforward. With distillate and bottoms mole fractions of the MVC specified, Nmin is easily calculated if αAB is known. If the relative volatility is not constant, αAB can be estimated from a geometric average as shown in Eq. (7-8b). This can be estimated for a first trial as
where αAB,R is determined from the bottoms composition and αAB,dist from the distillate composition.
Accurate use of the Fenske equation obviously requires an accurate value for the relative volatility. Smith (1963) covers in detail a method of calculating α by estimating temperatures and calculating the geometric average relative volatility. For approximate estimates this extra work is seldom necessary.
EXAMPLE 7-1. Fenske equation
A distillation column with a partial reboiler and a total condenser is separating a saturated vapor feed that is 40.0 mol% benzene (B), 30.0 mol% toluene (T), and 30.0 mol% cumene (C). Recovery of toluene in the distillate is 95%, and recovery of cumene in the bottoms is 98%. Reflux is a saturated liquid, and constant molal overflow (CMO) is valid. Pressure is at 1.0 atm. Relative volatilities are constant. Choosing toluene as the reference component, αB-T = 2.25 and αC-T = 0.21. Find the number of equilibrium stages required at total reflux, the recovery fraction of benzene in the distillate and in the bottoms, and the mole fractions in the distillate and bottoms.
Define. A total reflux column was shown in Figure 7-1. For T = toluene (LK), C = cumene (HK), B = benzene (LNK), we have αBT = 2.25, αTT = 1.0, αCT = 0.21, which means αTC = 1/0.21. zT = 0.3, zB = 0.4, zC = 0.3, FRLK,dist = FRT,dist = 0.95, and FRHK,bot = FRC,bot = 0.98.
Find N at total reflux.
Find FRB,dist at total reflux.
Find mole fractions of distillate and bottoms at total reflux.
Explore. Because operation is at total reflux and relative volatilities are constant, we can use the Fenske equation.
Plan. Calculate Nmin from Eq. (7-12) and then calculate FRB,dist from Eq. (7-13).
Do it.
Equation (7-12) gives
Note that αLK-HK = αtol-cumene = 1/αCumene-tol = 4.762.
Equation (7-13) gives
Benzene recovery in bottoms = 1 – FRB,dist = 0.0015. Note that
Dxi,dist = (FRi,dist)(Fzi) = 0.99805(0.4F) = 0.3994F.
D = Σ(Dxi,dist) = 0.9985(0.4F) + 0.95(0.3F) + (1 − 0.98)0.3F = 0.6904F.
Then, xBen,dist = 0.3994F/0.6904F = 0.5785, xTol,dist = 0.4128, xCum,dist = 0.0087.
B = F − D = 1 – 0.6904F = 0.3096F.
xBen,bot = (1 − 0.9985)(0.4F)/0.3096F = 0.001938, xTol,bot = 0.0485, xCum,Bot = 0.9496.
Check. The results can be checked by calculating FRC,dist using component A instead of B. The same answer is obtained.
Generalize. High recovery of a compound (e.g., the HK) in the bottoms means there will be very little of that compound in the distillate. Thus, the distillate is pure. To have high purity of the bottoms, we must have high recovery of the LK in the distillate.