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Separator Design 339

Example 6.7 Estimating the Number of Equilibrium Stages___________

This problem is adapted from a problem given by Fair and Bolles [56] for a de-
ethanizer column. A solution of hydrocarbons at its bubble point is pumped into
the column at an average pressure of 400 psia (27.6 bar). The composition of the
liquid feed is given in Table 6.7.1. Calculate the number of equilibrium stages if
the recovery of ethane in the top product is 99%, and the recovery of propylene in
the bottom product is also 99%. Also, determine the location of the feed point.
Follow the calculation procedure outlined in Table 6.28 using Equations
listed in Table 6.27. The light key (LK) is ethane and the heavy key (FIK) is pro-
pylene. First, calculate the composition of the top and bottom products. Then, de-
termine the optimum reflux ratio. Next, calculate the number of equilibrium
stages. Finally, calculate the location of the feed tray.
To obtain the composition of the top and bottom products, first calculate the
relative volatility of each component using the conditions of the feed as a first
guess. The relative volatility depends on temperature and pressure. The bubble
point of the feed at 400 psia (27.6 bar) and at the feed composition, calculated
using ASPEN [57], is 86.5 °F (130 °C). The K-values of the feed are listed in Ta-
ble 6.7.1. Bubble and dew points could also be calculated using K-values from the
DePriester charts [31] and by using the calculation procedures given in Chapter 3.
Next, calculate the relative volatility of the feed stream, defined by Equation
6.27.18, for each component relative to the heavy key component.
The relative volatility for each of the feed components in Table 6.7.1, will
now be used to calculate the composition at the top tray and the bottom product.
First calculate the constants AC and B c in Equation 6.27.1. Selecting one mole of
feed as the basis of calculation, the moles of ethane and propylene in the distillate
and bottom products, using the specified recoveries, are calculated as follows:
n; B = 0.99 (0.15) = 0.1485, moles of propylene in the bottom product

n; D = 0.01 (0.15) = 0.0015, moles of propylene in the top product

n iB = 0.01 (0.35) = 0.0035, moles of ethane in the bottom product

n iD = 0.99 (0.35) = 0. 3465, moles of ethane in the top product

Substituting into Equation 6.27.1 for propylene,

log (0.0015 / 0.1485) = AC + B c log 1.0

and for ethane,

log (0.3465 / 0.0035) = A c + B c log 2.238

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