Page 168 - Separation process engineering

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.
C. 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
V = L + D
j+1
j
y V = L x +Dx D
j j
j+1 j+1
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 = 0.0 (contains no methanol). Since steam is
s
saturated, S = V and B = L (constant molal overflow).

Then

(4-42)

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