Define. It helps to draw a schematic diagram of the apparatus, particularly since a new type of distillation is involved. This is shown in Figure 4-15. We wish to find the optimum feed plate location, N_F, and the total number of equilibrium stages, N, required for this separation. We could also calculate Q_c, D, B, and the steam rate S, but these were not asked for. We assume that the column is adiabatic since it is well insulated.

Explore. The first thing we need is equilibrium data. Fortunately, these are readily available (see Table 2-7 in Problem 2.D1).

Second, we would like to assume CMO so that we can use the McCabe-Thiele analysis procedure. An easy way to check this assumption is to compare the latent heats of vaporization per mole (Himmelblau, 1974).

These values are not equal, and in fact, water’s latent heat is 15.3% higher than methanol’s. Thus, CMO is not strictly valid; however, we will solve this problem assuming CMO and will check our results with a process simulator.

A look at Figure 4-15 shows that the configuration at the bottom of the column is different than when a reboiler is present. Thus we should expect that the bottom operating equations will be different from those derived previously.

Plan. We will use a McCabe-Thiele analysis. Plot the equilibrium data on a y-x graph.

Top Operating Line: Mass balances in the rectifying section (see Fig. 4-15) are

Assume CMO and solve for y_j+1.

Since the reflux is returned as a saturated liquid,

Enough information is available to plot the top operating line.

Feed Line:

Intersection: y = x = z

Once we substitute in values, we can plot the feed line.

Bottom Operating Line: The mass balances are

Solve for y:

Simplifications: Since the steam is pure water vapor, y_s = 0.0 (contains no methanol). Since steam is saturated, S = V and B = L (constant molal overflow).

Then

Equation 4-42

Note this is different from the operating equation for the bottom section when a reboiler is present. Slope = / (unknown), y intercept = −(/)x_B (unknown),

One known point is the intercept of the top operating line with the feed line. We still need a second point, and we can find it at the x intercept. When y is set to zero, x = x_B (this is left as Problem 4.C1).

Do It. Equilibrium data are plotted on Figure 4-16.

Top Operating Line:

We can plot this straight line as shown in Figure 4-16.

Feed Line: Slope = q/(q − 1) = 0.7/(0.7 − 1) = − 7/3.

Intersects at y = x = z = 0.6. Plotted in Figure 4-16.

Bottom Operating Line: We can plot this line between two points, the intercept of top operating line and feed line, and

This is also shown in Figure 4-16.

Step off stages, starting at the top. x₁ is in equilibrium with y₁ at x_D. Drawing a horizontal line to the equilibrium curve gives value x₁. y₂ and x₁ are related by the operating line. At constant y₂ (horizontal line), go to the equilibrium curve to find x₂. Continue this stage-by-stage procedure.

Optimum feed stage is determined as in Figure 4-8A. Optimum feed in Figure 4-16 is on stage 3 or 4 (since by accident x₃ is at intersection point of feed and operating lines). Since the feed is a two-phase feed, we would introduce it above stage 4 in this case.

Number of stages: Five is more than enough. We can calculate a fractional number of stages.

Equation 4-43

In Figure 4-16,

We need 4 + 0.9 = 4.9 equilibrium contacts.

Check. There are a series of internal consistency checks that can be made. Equilibrium should be a smooth curve. This will pick up misplotted points. L/V < 1 (otherwise no distillate product), and / > 1 (otherwise no bottoms product). The feed line’s slope is in the correct direction for a two-phase feed. A final check on the assumption of CMO would be advisable since the latent heats vary by 15%.

This problem was also run on the Aspen Plus process simulator (see Problem 4.G1 and chapter appendix). Aspen Plus does not assume CMO and with an appropriate vapor-liquid equilibrium (VLE) correlation (the nonrandom two-liquid model was used) should be more accurate than the McCabe-Thiele diagram, which assumes CMO. With 5 equilibrium stages and feed on stage 4 (the optimum location), x_D = 0.9335 and x_B = 0.08365, which doesn’t meet the specifications. With 6 equilibrium stages and feed on stage 5 (the optimum), x_D = 0.9646 and x_B = 0.0768, which is slightly better than the specifications. The differences in the McCabe-Thiele and process simulation results are due to the error involved in assuming CMO and, to a lesser extent, differences in equilibrium. Note that the McCabe-Thiele diagram is useful since it visually shows the effect of using open steam heating.

Generalize. Note that the y = x line is not always useful. Don’t memorize locations of points. Learn to derive what is needed. The total condenser does not change compositions and is not counted as an equilibrium stage. The total condenser appears in Figure 4-16 as the single point y = x = x_D. Think about why this is true. In general, all inputs to the column can change flow rates and hence slopes inside the column. The purpose of the feed line is to help determine this effect. The reflux stream and open steam are also inputs to the column. If they are not saturated streams the flow rates are calculated differently; this is discussed later.

Note that the open steam can be treated as a feed with q = (L_below − L_above)/S = (B—)/S = 0. Thus, B = . The slope of this feed line is q/(q − 1) = 0 and it intersects the y = x line at y = x = z = 0, which means the feed line for saturated steam is the x-axis.