Question

Exercise 6. Estimate the K-values and a of a vapour-liquid mixture of water (W) and methane (M) at P = 2 atm, T = 20 and 80°C. What is the effect of T on the component distribution? At these conditions, water exists mainly in the liquid phase and will follow Raoult's law. Because methane has a critical temperature of -82.5 °C, well below the temperatures of interest, it will exist mainly in the and follow Henry's law. Table 1: From Perry's Chemical Engineers’ Handbook, 6th ed. pp. 3-237 and 3-103 T[°C] 20 Pwater (atm) | 0.02307 0.4673 HCH. [atm] 3.76.104 6.82 · 104 80

Answer 1

his Vapouie - lauld equilibrium moro Ni lini 2 At pag arm , T = 25 c m Pwater Cater) : 0.02307 Hery latin) = 2.768104 +ạto cu to s} . PAHA 2 Petty, Hit Merty (20.02307) aum - 2.76x16" alom X Y CHA Mehu 2 3.26x103 2-0-02307 0-988 - 2 18783. 27 0.988 5.26x10h nely fore P=2atm ; T280°C Pinater (aten) 20-4673 Helm (exten) = 6.82x104 PRHY & HAM CHY (2-0.4473) = 6.82 * 10 Anche 2.24 X103 = Menu " Chy? Peter 2-0. 1673 20-7664 -0.4673 20 76.64 ka ya 1CH 0.7664 2 34214.3 Mely 2.24x103

x =P Vapour Drees SUSID 7 - Partial PA forespurte POL Tolo Press YAPTIVA). Y (1) (Piata (TYA) NAFX all-4) L = PALIWA) BAA & at 200 22 Yline Cray) a os soc | 0 1 1 1 1 (-y) 0.988 (-5211105). 1565222- 5.26*10 (1-0.985) 6 3 6 6 1 - 1 - 24 x 16) 2:24 *10 (1-0-26647 =1464 65.26 High vole of a shows that Hy will recomain in gos phose, here & gits reclative volatility. Effect of T on component diskebution- methane has a critical temp. of - 82.56 so, methane cill reemain gos Phose always above this teme so in this case component distribution will charge only because of coatere Vapour foremed on tncreeon the temp.