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

# 02 ya f 2 2 nf np2 16 the final equations are 6

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Unformatted text preview: F 2 2 nF nP2 (16) The final equations are (6) combined with (12) and (14) through (16). (c) From Eq. (1-3), yA P nP1 1 yA P nP2 =9 (17) 2 and, yBP nP1 1 yBP nP2 2 = 1 9 (18) Combining (17) and (18), with component and total material balances around the separator gives the following equations that can be used with Eq. (6): yA P = 0.9 yA F 1 nF nP1 (19) Exercise 2.9 (continued) Analysis: (c) (continued) yA P = 01yA F . 2 nF nP2 (20) nP1 = nF 0.8 yA F + 01 . (21) nP2 = nF 0.9 − 0.8 yA F (22) (d) From Eq. (1-5), SPA,B = SFA / SFB = 361 1 − SFA / 1 − SFB (23) SFA = 0.95 is given. Combining this with Eq. (23), gives SFB = 0.05. This part then proceeds as in part (b) to give: nP1 = nF 0.90 yA F + 0.05 (24) nP2 = nF − nP1 or nP2 = nF 0.95 − 0.90 yA F (25) yA P = 0.95 yA F 1 yA P = 0.05 yA F 2 nF nP2 nF nP1 (26) (27) Equations (24) through (27) are combined with Eq. (6). A spreadsheet can be used to compute the dimensionless minimum work, Wmin RT0nF with the following results for Parts (a) through (d): Exercise 2.9 (continued) From the plot on the previous page, it is seen that the dimensionless minimum work is very sensitive to the feed mole fraction and to the product purities. From the derivation of the minimum work equations, it is seen that they are independent of the separation method and only depend on thermodynamics. To prove that the largest value of Wmin occurs for a feed with equimolar quantities of A and B, consider the case of a perfect separation, part (a), as given by Eq. (11). Let W = Wmin/nFRT0 and y = yA F . Then, the derivative of W with respect to y is, dW y −1 = − ln y + − ln(1 − y ) + (1 − y ) dy y 1− y = − [ ln y − ln(1 − y ) ] For min/max, set the derivative to zero and solve for y. Therefore, dW = 0 = − [ ln y − ln(1 − y ) ] dy Solving, y = 1 − y or y = 0.5. This is an equimolar feed. Furthermore, it gives a maximum value of W . Exercise 2.10 Subject: Relative volatility of the isopentane-normal pentane system Given: Experimental data for relative volatility 125-250oF. Vapor pressure constants. Find: Relative volatilities from Raoult's law over the same temperature range and compare them to the experimental values. Analysis: Combining Eq. (5), Table 2.3, for Raoult's law, with Eq. (2-21) for α . s PiC α iC5 ,nC5 = s 5 (1) PnC5 Only the first three constants of the extended Antoine equation are given. (T=K, P= kPa) 2345.09 s PiC5 = exp 13.6106 − T − 40.2128 (2), (3) 2554.60 s PnC5 = exp 13.9778 − T − 36.2529 Using a spreadsheet to calculate the relative volatility from Eqs. (1), (2), and (3): s T, F T, K Raoult's law Expt. α PiC5 , kPa PnsC5 , kPa α 125 1.26 325 214 167 1.28 150 1.23 339 314 251 1.25 352 446 364 1.22 175 1.21 200 1.18 366 614 512 1.20 225 1.16 380 823 700 1.18 250 1.14 394 1079 834 1.16 The Raoult's law values are within 2% of the experimental values. Exercise 2.11 Subject: Condenser duty of a vacuum distillation column separating ethyl benzene (EB) and styrene (S). Given: Phase condition, temperature, pressure, flow rate, and compositions for streams entering and leaving a condenser, which produces subcoole...
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