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(a) Adsorbent flow rate per 1,000 m3 of gas if 1.2 times the minimum flow rate is used.
(b) Number of theoretical stages needed.
First compute the material balance on the gas, taking a basis of one hour. From
the ideal gas law, the gas feed rate is Pυ (1)(1,000)(106 )
= 40,900 mol / h of gas
This gas contains 0.45(40,900) = 18,400 mol/h C3= and 40,900 - 18,400 = 22,500 mol/h C3.
Assume a configuration analogous to that for liquid-liquid extraction with reflux, as shown in
Fig. 8.26b. Entering solvent, SB, becomes the adsorbent: raffinate, R, becomes the propane-rich
product; Stream D + LR is desorbed from the adsorbent in the desorber (analogous to the Solvent
removal step in extraction), with D becoming the propylene-rich product and LR, becoming the
Extract reflux. The upward movement of adsorbent and adsorbate is V and the downward
movement of non-adsorbed gas is L. Other configurations are possible, but the one just
described is used here.
From the experimental equilibrium data of Exercise 15.9, assume that at 1atm, the total
adsorbent loading remains constant at a value of 2.0 mmol/g = 2.0 mol of combined C3 and C3=
adsorbed at equilibrium per kg of adsorbent. Also, in Exercise 15.9, although the mixture data
scatter, a relative selectivity = αC3=,C3 = (yC3=/xC3=)/(yC3/xC3) = approximately 2.5, with mole
fractions y in the adsorbate an mole fractions x in the non-adsorbed gas. With these two
assumptions, a relatively simple McCabe-Thiele method is used to solve this exercise.
Because the flow rate of adsorbent is constant throughout the stages, both below and
above the gas feed entry, let V = adsorbate (adsorbent-free) flow rate. Then, because of the
assumption of constant loading, V is constant throughout the stages. The non-adsorbed gas flow
rate, L, is constant above the feed at a flow rate of LR;. below the feed, it is R. Thus, on a
McCabe-Thiele diagram, the operating lines have constant slopes. By a C3 material balance
around the system, followed by a total material balance, Fx FC3 = 40,900(0.55) = 22,500 = y DC3 D + x BC3 B = 010 D + 0.90 B
. (1) F = 40,900 = D + B (2) Solving Eqs (1) and (2), D = 17,900 mol/h and B = 23,000 mol/h Exercise 15.34 (continued)
The McCabe-Thiele diagram is shown on the following page in mole fractions of C3=,
where the equilibrium curve is computed from Eq. (7-3), y = αx/[1+x(α−1)]. The q-line is
vertical, because the adsorbent that passes over the feed stage is already loaded with adsorbate, so
that no change in the adsorbate flow rate, V, occurs. To determine the minimum adsorbent flow
rate, the usual minimum reflux construction, shown below, is made (corresponding to an infinite
number of stages), with an operating line that starts at y = x = 0.90 and terminating at the
intersection with the equilibrium line and the q-line x = 0.45 and x = 0.45 and y = 0.6716 from
the equilibrium equation. Thus, the slope of this operating line = (0.90-0.6716)/(0.9-0.45) =
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