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

12 continued analysis continued 2 2 0 0 1 0 05 1 1 b2

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Unformatted text preview: file, for the tear variables and the total liquid flow rates are: Stage j 1 (Total condenser) 2 3 4 5 (Partial reboiler) Vj , kmol/h 0 2600 2600 2600 2600 Tj , oC 64.7 70.6 76.4 82.3 88.1 Lj , kmol/h 2000 2000 3000 3000 400 Compute K-values at 1 atm and the above temperatures for each component by Eq. (2-44), K = Pis/P. From the 7th edition of Perry's Handbook, on page 2-50 to 2-54, in Table 2-6, vapor pressure data are fitted to an extended Clausius-Clapeyron equation: P s = exp C1 + C2 + C 3 ln T + C 4T C 5 , P s is in Pa and T is in K , where the constants are: T Component M E P C1 81.768 74.475 88.134 C2 -6876 -7164.3 -8498.6 C3 -8.7078 -7.327 -9.0766 C4 7.1926E-6 3.1340E-6 8.3303E-18 C5 2 2 6 Exercise 10.11 (continued) Analysis: (continued) Computed K-values at the five assumed stage temperatures for P = 101300 Pa are: Stage Temperature, K K-values: M E P 1 337.9 2 343.7 3 349.6 4 355.4 5 361.3 1.0086 0.5705 0.2505 1.2639 0.7309 0.3294 1.5702 0.9272 0.4281 1.9350 1.1654 0.5504 2.3665 1.4521 0.7002 A plot of the K-values as a function of temperature is shown on the next page. The component material balance equations, combined with the equilibrium equations to eliminate vapor-phase mole fractions, for each of the 5 stages are as follows: Stage 1: − D + L1 xi ,1 + V2 Ki ,2 xi ,2 = 0 − 2600 xi ,1 + 2600 Ki ,2 xi ,2 = 0 Stage 2: L1 xi ,1 − V2 Ki ,2 + L2 xi ,2 + V3 Ki ,3 xi ,3 = 0 2000 xi ,1 − 2600 Ki ,2 + 2000 xi ,2 + 2600 Ki ,3 xi ,3 = 0 Stage 3: (3) L3 xi ,3 − V4 Ki ,4 + L4 xi ,4 + V5 Ki ,5 xi ,5 = 0 3000 xi ,3 − 2600 Ki ,4 + 3000 xi ,4 + 2600 Ki ,5 xi ,5 = 0 Stage 5: (2) L2 xi ,2 − V3 Ki ,3 + L3 xi ,3 + V4 Ki ,4 xi ,4 = − F3 zi ,3 2000 xi ,2 − 2600 Ki ,3 + 2000 xi ,3 + 2600 Ki ,4 xi ,4 = −1000zi ,3 Stage 4: (1) (4) L4 xi ,4 − V5 Ki ,5 + B xi ,5 = 0 3000 xi ,4 − 2600 Ki ,5 + 400 xi ,5 = 0 (5) For each component, the K-values for each stage are substituted from the table above. The resulting equations form 3 tridiagonal matrix equations, one each for each of the 3 components. These equations are as follows, as produced from a spreadsheet: Exercise 10.11 (continued) Analysis: (continued) Analysis: (continued) Exercise 10.11 (continued) ___________________________________________________________________________________________ Exercise 10.11 (continued) Analysis: (continued) A new set of stage temperatures is computed by running bubble-point temperature calculations to find a temperature that satisfies Eq. (10-21), using the normalized mole fractions with the vapor pressure equations and a pressure of 1 atm. The results are as follows compared to the initially assumed values: Stage j 1 (Total condenser) 2 3 4 5 (Partial reboiler) Tj , o C Tj , o C initially from 1st iteration assumed 64.7 67.2 70.6 69.6 76.4 73.1 82.3 77.8 88.1 84.1 Exercise 10.12 Subject: Solving a block tridiagonal matrix equation. Given: 9 linear equations in 9 unknowns. Find: Values of the 9 unknowns using the block form of the Thomas algorithm. Analysis: To form...
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