Page 133 - Separation process engineering

(3-4)

Don’t memorize equations like these; they can be derived as needed.

For the energy balance we will use the convention that all heat loads will be treated as inputs. If energy is
removed, then the numerical value of the heat load will be negative. The steady-state energy balance
around the entire column is

(3-5)
where we have assumed that kinetic and potential energy and work terms are negligible. The column is
assumed to be well insulated and adiabatic. Q will be positive and Q negative. The enthalpies in Eq.
c
R
(3-5) can all be determined from an enthalpy-composition diagram (e.g., Figure 2-4) or from the heat
capacities and latent heats of vaporization. In general,

(3-6a, b, c)

These three enthalpies can all be determined. We would find h on Figure 2-4 on the saturated liquid
B
(boiling-line) at x = x .
B
Since F was specified and D and B were just calculated, we are left with two unknowns, Q and Q , in
R
c
Eq. (3-5). Obviously another equation is required.
For the total condenser shown in Figure 3-8 we can determine Q . The total condenser changes the phase
c
of the entering vapor stream but does not affect the composition. The splitter after the condenser changes
only flow rates. Thus composition is unchanged and

(3-7)

The condenser mass balance is

(3-8)

Since the external reflux ratio, L /D, is specified, we can substitute its value into Eq. (3-8).
0

(3-9)

Then, since the terms on the right-hand side of Eq. (3-9) are known, we can calculate V . The condenser
1
energy balance is

(3-10)

Since stream V is a vapor leaving an equilibrium stage in the distillation column, it is a saturated vapor.
1
Thus,