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Separation Process Principles- 2n - Seader & Henley - Solutions Manual

00723219 2318 eqmin 36935 ideal loading time to

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Unformatted text preview: , (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. Analysis: 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 ) F= = = 40,900 mol / h of gas RT (82.06)(298) 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) Analysis: (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) = 0.5706...
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