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

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online