Picture Of Curved Operating Lines Mccabe-thiele

Mccabe Thiele Tray Efficiency

Figure 31. McCabe-Thiele analysis of equilibrium stages for the ethanol-water system.

x (ethanol)

Figure 31. McCabe-Thiele analysis of equilibrium stages for the ethanol-water system.

the slope of this line may be determined:

Slope =

The operating line for the top of the column (see Figure 31) is then y = (0.56)x + (1 - 0.56)(0.7) = 0.56x + 0.308

The describing equation for the operating line at the bottom of the column is

Substitution yields

V 1.25 (0.2 - 0.05) + 1.2 (0.7 - 0.05) - (0.7 - 0.2) = 2.07

The operating line for the bottom of the column (see Figure 31) is y = (2.07)x - (2.07 - 1)(0.2) = 2.07* - 0.0535


By counting the number of equilibrium stages in Figure 31, one obtains five equilibrium stages. Note, however, that the partial reboiler counts as a stage. The number of equilibrium stages not counting the reboiler, N, is four. The stage efficiency E0 is used to find the actual number of stages:

% 6 actual stages

The column height is

The feed line is drawn from the x = 0.2 on the y = x line through the intersection of the top and bottom operating lines to the equilibrium curve (see Figure 31). The optimum feed stage from the diagram is equilibrium stage 4, which is the lowest stage above the reboiler. This corresponds to the sixth actual stage. The optimum feed stage is stage 6.

Finally, the vapor volumetric flowrate is calculated:

_/2501bmol\/ lh \ /379 ft3\ /300 + 460°R\ CL ~~ \ h J V 3600 s J \ lbmol J { 60 + 460 R )

This enables one to estimate the diameter at the top of the column:

q 38.47 ft3/s „ _ „2 Area = - = —rc , ' = 7.693ft2 v 5 ft/s


J.B. Chemicals, Inc., has hired Theodore Distillation Consultants (TDC) to assist in the separation of 500 kmol/h of a saturated vapor feed (quality, q = 0) consisting of 40 mol % benzene (C), 30 mol % toluene (A), and 30 mol % cumene (B). It has been suggested that the distillation column include a partial reboiler and a total condenser. J.B. Chemicals requires 99% recovery of the toluene in the distillate (FRA)dist and 95% recovery of the cumene in the bottoms (FRB)bot. As an engineer for TDC you need to determine the actual number of equilibrium stages required for the separation, as well as the fractional recovery (FR) of the benzene in the distillate. The column will operate at a reflux ratio of 2.0 times the minimum and will have relative volatilities of aCA = 2, aBA = 0.25, and aAA = 1.0.

A = toluene (light key) B = cumene (heavy key) C = benzene (light nonkey)


Fenske developed the following equation to calculate the minimum number of stages at total reflux for the separation of two key components (i.e., A and B in a mixture of A, B, and C). The original notation is employed here

where Nmin = minimum number of equilibrium stages

FR, = fractional recovery in distillate (dist) or bottoms (bot) for component i ocBA = relative volatility of component B versus that of component A

The following equation can be used to calculate the fractional recoveries of the nonkey components (i.e., of C) in the distillate and bottoms:



The Underwood equation requires a trial-and-error solution and a subsequent material balance to estimate the minimum reflux ratio. First, the unknown, q>, is determined by trial and error, such that both sides of the following equation are equal. The unknown value of <p should lie between the relative volatilities of the light and heavy key components. The key components are those that have their fractional recoveries specified. The most volatile component of the keys is the light key and the least volatile is the heavy key. All other components are referred to as nonkey components. If a nonkey component is lighter than the light key component, it is a light nonkey; if it is heavier than the heavy key component it is a heavy nonkey component.

where A I 'fied = change in vapor flowrate at feed stage F = feed rate q = percentage of feed remaining liquid zf = feed composition of component i n = number of components

<p = unknown root to be determined by trial and error Next, calculate the minimum vapor flowrate, Fmin, from the following equation:

where Dxi dist = amount of component i in the distillate = Fz;(FR),dist

Once ( min is known, the minimum liquid flowrate is calculated from the following:

The minimum reflux ratio is then calculated by dividing the minimum liquid flowrate by the distillate flowrate as shown below:

The Gilliland correlation uses the results of the Fenske and Underwood equations to determine the actual number of equilibrium stages. The correlation has been fit to three equations:

= 0.545827 - 0.591422x + 0.002743/x for 0.01 ^ x sj 0.90 = 0.16595 -0.16595x for 0.9 ^ 1.0

where N — actual number of equilibrium stages x = [(L/D)— (L/D)mm]/[(L/D) + 1]

The values of jVmin and (L/D)mjn have been previously defined as the minimum number of equilibrium stages (Fenske equation) and minimum reflux ratio (Underwood equation).

Thus, the calculational procedure is as follows:

1. Calculate the minimum number of equilibrium stages, Nmin, from the Fenske equation.

2. Calculate the minimum reflux ratio, (L/D)min, from the Underwood equation.

3. Select an actual reflux ratio, L/D, which usually ranges from 1 to 2.5 times the minimum ratio.

4. Calculate x.

5. Solve the Gilliland correlation for the actual number of equilbrium stages.

Following this procedure, the minimum number of equilibrium stages, A/rmjn, at total reflux is

The fractional recovery of benzene, FRC, in the distillate is

Then cp is determined by trial and error using the Underwood equation: AVteoi = F(l-q) = ± a'Fz<

_ (1)(500)(0.3) (0.25)(500)(0.3) (2)(500)(0.4) (1 )-<p + (0.25) — q> + (2) — (p cp = 0.561678

The recovery rate of each component in the distillate, Dxi, and the total distillate flowrate, D, are obtained from a simple material balance:

Toluene: ZAF(FRAdist) = DA (3)(500)(0.99) = 148.5 kmol/h Cumene: ZBF(FRB_dist) = D0 (0.3)(500)(1 - 0.95) = 7.5 kmol/h Benzene: ZCF(FRC dist) = Dc (0.4)(500)(0.9998) = 199.95 kmol/h

The minimum overhead vapor flow rate, Vmin is

1 -0.561678 0.25 -0.561678 2-0.561678 = 610.81 kmol/h

The minimum reflux ratio, (¿/D)min, is calculated from

The actual reflux ratio is

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