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

# 8 i 2 3zv zv z zv a b b b a 2 zv v a b b 2 zv

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Unformatted text preview: Exercise 10.14 Subject: Obtaining derivatives of K-values for use in the Jacobian of the simultaneouscorrection method. Given: Chao-Seader correlation for the K-value based on Eq. (2) in Table 2.3. Find: Analytical derivatives for: ∂Ki , j ∂Tj , ∂Ki , j ∂υ i ,k , and ∂Ki , j ∂li , k . where, υI,k = yi,kVk and l = xi,kLk and component molar flow rates. γφ Analysis: From Eq. (2) in Table 2.3, Ki = iL iL φiV From Eqs. (2-64) and (2-62), v γ iL = exp iL δ i − RT C j =1 (1) 2 Φ jδ j (2) From Eq. (2-61), where for a given component, i , viL and δ i are constants, Φi = C xi viL j =1 (3) x j viL It is important to distinguish between the symbols υi,k (the molar vapor component flow rate) and viL (the component molar volume in the liquid phase = a constant). For the liquid-phase, pure-component, fugacity coefficient, the Chao-Seader equations are: (0) (1) log φiL = log φiL + ωi log φiL (0) logφiL = A0 + ( (4) A1 + A2Tri + A3Tri2 + A4Tri3 + Pri A5 + A6Tri + A7Tri2 + Tri ( ) ) Pri2 A8 + A9Tri − log Pri (1) logφiL = A10 + A11Tri + where, Tr = T/Tc (5) A12 + A13Tri3 + A14 Pri − 0.6 Tri ( ) (6) Exercise 10.14 (continued) Analysis (continued) From Eqs. (2-56), (2-46), (2-47), (2-48), (2-49), and (2-50), Bi A AB B − ln ZV − B − 2 i − i ln 1 + B B AB ZV φiV = exp ZV − 1 3 2 ZV − ZV + ZV A − B − B 2 − AB = 0 A= aP R 2T 2 Derivative B= (9) ∂Ki , j ∂Tj bP RT (7) (8) (10) with a and b given Table 2.5 and Eqs. (2 - 49) and (2 - 50) : Using Eq. (1) with the chain rule, ∂γ iL From Eq. (2), =− ∂Tj ∂Ki , j ∂Tj = φiL ∂γ iL γ ∂φ γ φ ∂φ + iL iL − iL 2 iL iV φiV ∂Tj φiV ∂Tj φiV ∂Tj γ iL viL δ i − C j =1 2 Φ jδ j RTj2 (0 (1 ∂φiL ∂ log φ iL ) ∂ log φ iL) = 2.3026 φ iL + ωi ∂Tj ∂Tj ∂Tj where from Eqs. (5) and (6), (0 ∂ log φ iL ) 1 A1 = − 2 + A2 + 2 A3Tri + 3 A4 Tri2 + Pri A6 + 2 A7 Tri + Pri2 A9 ∂Tj Tci Tri (1 ∂ log φ iL) 1 A = A11 − 12 + 3 A13Tri2 ∂Tj Tci Tri2 To obtain ∂φiV , we note that in Eq. (7), ZV, A, Ai, B, and Bi are functions of T. ∂Tj Therefore, from Eq. (7), (11) Exercise 10.14 (continued) Analysis: (continued) Bi ∂ZV 1 ∂Bi Bi ∂B 1 + ZV − 1 −2 − B ∂T B ∂T B ∂T ZV − B ∂φiV = φiV ∂Tj 1 B A BZV where, ∂ZV ∂B − − ∂T ∂T Ai Bi ∂A A A 2 B ∂B 1 − − 2 2 i− i + A B ∂T B A B ∂T B ∂B B ∂ZV − ∂T ZV ∂T 1+ B ZV 2 A ∂Ai A ∂B B − 2 i ln 1 + − Ai ∂T B ∂T ZV Ai Bi − AB ∂A A ∂Ai A ∂B B = −2.5 , = −2.5 i , =− , ∂T T ∂T T ∂T T ∂Bi B =− i ∂T T ∂ZV is obtained as follows by implicit integration from Eq. (8): ∂T ∂ZV ∂Z ∂ZV ∂A ∂B ∂B ∂B ∂A − 2 ZV V + A − B − B 2 + ZV − − 2B −A −B =0 ∂T ∂T ∂T ∂T ∂T ∂T ∂T ∂T Solving, 2 3ZV ∂ZV = ∂T ZV 2 ∂A ∂B ∂B ∂B ∂A Z −2.5 A + B + 2 B + 35 AB . − − 2B −A −B V TT T T ∂T ∂T ∂T T ∂T = 2 2 2 ZV − 3ZV − A − B − B 2 2 ZV − 3ZV − A − B − B 2 Substitution of the equations below Eq. (11) into Eq. (1...
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